General Solution Help Please (1 Viewer)

Midori-Days · Aug 11, 2011

Question:

Find the general solution of the equation:
tan(theta) = sin2(theta)

I don't know how to go about with a Q like this :/

michaeljennings · Aug 11, 2011

just guessing but i think you would rewrite it as sintheta over costheta equals 2sinthetacostheta and solve from there

Then you get sin theta =0 and cos theta equals to plus minus 1 over square root 2

Drongoski · Aug 11, 2011

Midori-Days said:
Question:

Find the general solution of the equation:
tan(theta) = sin2(theta)

I don't know how to go about with a Q like this :/

$sin \theta = sin 2 \theta = 2 sin \theta cos\theta$

$\therefore sin \theta (2cos \theta -1) = 0$

$\therefore either: sin \theta = 0 \implies \theta = \pm n \pi (n = 0, 1, 2, 3, . . .)$

or
$2cos \theta - 1 = 0 \implies cos \theta = \frac {1}{2} \implies \theta = \pm 2n \pi \pm \frac {\pi}{3} (n = 0, 1,2, 3 . . . .)$

Midori-Days · Aug 11, 2011

Drongoski said:
$sin \theta = sin 2 \theta = 2 sin \theta cos\theta$

$\therefore sin \theta (2cos \theta -1) = 0$

$\therefore either: sin \theta = 0$

or
$2cos \theta - 1 = 0 \implies cos \theta = \frac {1}{2}$

$sin \theta /cos\theta = 2sin\theta cos\theta$

$1 /cos\theta = 2cos\theta$

$1 = 2cos^2\theta$

$1 /2 = cos^2\theta$

$1 = cos2\theta + 1$

$0 = cos2\theta$

can u do that? O_O

michaeljennings · Aug 11, 2011

Midori-Days said:
$sin \theta /cos\theta = 2sin\theta cos\theta$

$1 /cos\theta = 2cos\theta$

$1 = 2cos^2\theta$

$1 /2 = cos^2\theta$

$1 = cos2\theta + 1$

$0 = cos2\theta$

can u do that? O_O

You cant throw away the sintheta though you have to include its general solution of 0 in your answer as well

Midori-Days · Aug 11, 2011

okay thanks...then i'll use what Drongoski stated above

michaeljennings · Aug 11, 2011

good luck for your trials

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General Solution Help Please (1 Viewer)

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